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Unit used in continuous time process noise matrix in kalman filters, when STD is from discrete time data

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0

1

$begingroup$

I'm trying to make a process noise matrix in continuous time. But i can't seem to find a clear definition of what "unit" the matrix should contain in continuous time.

From our control book we have $V_2d=V_2c/T$, where $V_2$ is our measurement noise matrix. From the web i found the conversion for discrete time standard deviation to continuous time: $V_1c=V_1d/Ts$, where $V_1$ is process noise. Should this contain the variance, noise power or standard deviation? However isn't this different from the conversion of $V_2$, why is this?

The Band-Limited White Noise block from simulink/matlab takes in a noise power, $N_p = sigma^2cdot Ts$. So the question is... why specifically noise power? Normally power is the energy over time, is this also the case for noise power, if yes, isn't noise energy the same as variance?

If you have some sources for the information about it, i would happily look through it. It is very confusing to work with it without clear definitions.

discrete-mathematics stochastic-processes stochastic-analysis kalman-filter noise

share|cite|improve this question

asked Mar 21 at 15:08

CLover32CLover32

1

$endgroup$

add a comment | 

0

1

$begingroup$

I'm trying to make a process noise matrix in continuous time. But i can't seem to find a clear definition of what "unit" the matrix should contain in continuous time.

From our control book we have $V_2d=V_2c/T$, where $V_2$ is our measurement noise matrix. From the web i found the conversion for discrete time standard deviation to continuous time: $V_1c=V_1d/Ts$, where $V_1$ is process noise. Should this contain the variance, noise power or standard deviation? However isn't this different from the conversion of $V_2$, why is this?

The Band-Limited White Noise block from simulink/matlab takes in a noise power, $N_p = sigma^2cdot Ts$. So the question is... why specifically noise power? Normally power is the energy over time, is this also the case for noise power, if yes, isn't noise energy the same as variance?

If you have some sources for the information about it, i would happily look through it. It is very confusing to work with it without clear definitions.

discrete-mathematics stochastic-processes stochastic-analysis kalman-filter noise

share|cite|improve this question

asked Mar 21 at 15:08

CLover32CLover32

1

$endgroup$

add a comment | 

$begingroup$

I'm trying to make a process noise matrix in continuous time. But i can't seem to find a clear definition of what "unit" the matrix should contain in continuous time.

From our control book we have $V_2d=V_2c/T$, where $V_2$ is our measurement noise matrix. From the web i found the conversion for discrete time standard deviation to continuous time: $V_1c=V_1d/Ts$, where $V_1$ is process noise. Should this contain the variance, noise power or standard deviation? However isn't this different from the conversion of $V_2$, why is this?

The Band-Limited White Noise block from simulink/matlab takes in a noise power, $N_p = sigma^2cdot Ts$. So the question is... why specifically noise power? Normally power is the energy over time, is this also the case for noise power, if yes, isn't noise energy the same as variance?

If you have some sources for the information about it, i would happily look through it. It is very confusing to work with it without clear definitions.

discrete-mathematics stochastic-processes stochastic-analysis kalman-filter noise

share|cite|improve this question

asked Mar 21 at 15:08

CLover32CLover32

1

$endgroup$

share|cite|improve this question

asked Mar 21 at 15:08

CLover32CLover32

1

share|cite|improve this question

asked Mar 21 at 15:08

CLover32CLover32

1

asked Mar 21 at 15:08

CLover32CLover32

1

asked Mar 21 at 15:08

CLover32CLover32

1

1 Answer
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0

$begingroup$

The reverse, so from continuous time to discrete time, can be achieved using (with zero order hold for $u(t)$)

beginalign
A_d &= e^A_c,T, \
B_d &= int_0^T e^A_c,tau dtau,B_c, \
C_d &= C_c, \
D_d &= D_c, \
W_d &= int_0^T e^A_c,tau,W_c,e^A_c^top,tau dtau, \
V_d &= V_c,T,
endalign

where the subscripts $c$ and $d$ stand for the continuous and discrete time state space models respectively and $T$ the discretization time step size. The continuous time model would be

beginalign
dotx(t) &= A_c,x(t) + B_c,u(t) + w(t), quad w(t) sim mathcalN(0,W_c), \
y(t) &= C_c,x(t) + D_c,u(t) + v(t), quad v(t) sim mathcalN(0,V_c),
endalign

such that the discrete time model becomes

beginalign
x_k+1 &= A_d,x_k + B_d,u_k + w_k, quad w_k sim mathcalN(0,W_d), \
y_k &= C_d,x_k + D_d,u_k + v_k, quad v_k sim mathcalN(0,V_d).
endalign

When all matrices are constant during the discretization time step the expression for $W_d$ can be simplified to the following impliciet equation

$$
A_c,W_d + W_d,A_c^top = A_d,W_c,A_d^top - W_c.
$$

So if $A_d$, $A_c$ and $W_d$ are known the above equation is just a discrete Lyapunov equation in $W_c$.

share|cite|improve this answer

answered Mar 23 at 16:19

Kwin van der VeenKwin van der Veen

5,6502828

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Our book Linear Systems Control, DETERMINISTIC AND STOCHASTIC METHODS published by Springer says that the measurement noise in discrete time is defined as $V_2d = V2/T$. To me it makes sense, since you need to increase the intensity when decreasing the sampling time. Is the book just wrong or?
  $endgroup$
  – CLover32
  Mar 24 at 14:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @CLover32 I just quoted what is stated on the Wikipedia page, so it can very well be wrong. Also the main focus of my answer was on the relation between the process noise $W_c$ and $W_d$. However I did notice a mistake in the book, namely equation (6.289) on page 414 says $e^A,tau=1-A,tau+frac12!A^2,tau^2-cdots$, but it should be $e^A,tau=I+A,tau+frac12!A^2,tau^2+cdots$ so I can't exclude the possibility that the book might be wrong as well.
  $endgroup$
  – Kwin van der Veen
  Mar 24 at 15:06
  

  
  
  
  

add a comment | 

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0

$begingroup$

The reverse, so from continuous time to discrete time, can be achieved using (with zero order hold for $u(t)$)

beginalign
A_d &= e^A_c,T, \
B_d &= int_0^T e^A_c,tau dtau,B_c, \
C_d &= C_c, \
D_d &= D_c, \
W_d &= int_0^T e^A_c,tau,W_c,e^A_c^top,tau dtau, \
V_d &= V_c,T,
endalign

where the subscripts $c$ and $d$ stand for the continuous and discrete time state space models respectively and $T$ the discretization time step size. The continuous time model would be

beginalign
dotx(t) &= A_c,x(t) + B_c,u(t) + w(t), quad w(t) sim mathcalN(0,W_c), \
y(t) &= C_c,x(t) + D_c,u(t) + v(t), quad v(t) sim mathcalN(0,V_c),
endalign

such that the discrete time model becomes

beginalign
x_k+1 &= A_d,x_k + B_d,u_k + w_k, quad w_k sim mathcalN(0,W_d), \
y_k &= C_d,x_k + D_d,u_k + v_k, quad v_k sim mathcalN(0,V_d).
endalign

When all matrices are constant during the discretization time step the expression for $W_d$ can be simplified to the following impliciet equation

$$
A_c,W_d + W_d,A_c^top = A_d,W_c,A_d^top - W_c.
$$

So if $A_d$, $A_c$ and $W_d$ are known the above equation is just a discrete Lyapunov equation in $W_c$.

share|cite|improve this answer

answered Mar 23 at 16:19

Kwin van der VeenKwin van der Veen

5,6502828

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Our book Linear Systems Control, DETERMINISTIC AND STOCHASTIC METHODS published by Springer says that the measurement noise in discrete time is defined as $V_2d = V2/T$. To me it makes sense, since you need to increase the intensity when decreasing the sampling time. Is the book just wrong or?
  $endgroup$
  – CLover32
  Mar 24 at 14:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @CLover32 I just quoted what is stated on the Wikipedia page, so it can very well be wrong. Also the main focus of my answer was on the relation between the process noise $W_c$ and $W_d$. However I did notice a mistake in the book, namely equation (6.289) on page 414 says $e^A,tau=1-A,tau+frac12!A^2,tau^2-cdots$, but it should be $e^A,tau=I+A,tau+frac12!A^2,tau^2+cdots$ so I can't exclude the possibility that the book might be wrong as well.
  $endgroup$
  – Kwin van der Veen
  Mar 24 at 15:06
  

  
  
  
  

add a comment | 

0

$begingroup$

The reverse, so from continuous time to discrete time, can be achieved using (with zero order hold for $u(t)$)

beginalign
A_d &= e^A_c,T, \
B_d &= int_0^T e^A_c,tau dtau,B_c, \
C_d &= C_c, \
D_d &= D_c, \
W_d &= int_0^T e^A_c,tau,W_c,e^A_c^top,tau dtau, \
V_d &= V_c,T,
endalign

where the subscripts $c$ and $d$ stand for the continuous and discrete time state space models respectively and $T$ the discretization time step size. The continuous time model would be

beginalign
dotx(t) &= A_c,x(t) + B_c,u(t) + w(t), quad w(t) sim mathcalN(0,W_c), \
y(t) &= C_c,x(t) + D_c,u(t) + v(t), quad v(t) sim mathcalN(0,V_c),
endalign

such that the discrete time model becomes

beginalign
x_k+1 &= A_d,x_k + B_d,u_k + w_k, quad w_k sim mathcalN(0,W_d), \
y_k &= C_d,x_k + D_d,u_k + v_k, quad v_k sim mathcalN(0,V_d).
endalign

When all matrices are constant during the discretization time step the expression for $W_d$ can be simplified to the following impliciet equation

$$
A_c,W_d + W_d,A_c^top = A_d,W_c,A_d^top - W_c.
$$

So if $A_d$, $A_c$ and $W_d$ are known the above equation is just a discrete Lyapunov equation in $W_c$.

share|cite|improve this answer

answered Mar 23 at 16:19

Kwin van der VeenKwin van der Veen

5,6502828

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Our book Linear Systems Control, DETERMINISTIC AND STOCHASTIC METHODS published by Springer says that the measurement noise in discrete time is defined as $V_2d = V2/T$. To me it makes sense, since you need to increase the intensity when decreasing the sampling time. Is the book just wrong or?
  $endgroup$
  – CLover32
  Mar 24 at 14:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @CLover32 I just quoted what is stated on the Wikipedia page, so it can very well be wrong. Also the main focus of my answer was on the relation between the process noise $W_c$ and $W_d$. However I did notice a mistake in the book, namely equation (6.289) on page 414 says $e^A,tau=1-A,tau+frac12!A^2,tau^2-cdots$, but it should be $e^A,tau=I+A,tau+frac12!A^2,tau^2+cdots$ so I can't exclude the possibility that the book might be wrong as well.
  $endgroup$
  – Kwin van der Veen
  Mar 24 at 15:06
  

  
  
  
  

add a comment | 

$begingroup$

The reverse, so from continuous time to discrete time, can be achieved using (with zero order hold for $u(t)$)

beginalign
A_d &= e^A_c,T, \
B_d &= int_0^T e^A_c,tau dtau,B_c, \
C_d &= C_c, \
D_d &= D_c, \
W_d &= int_0^T e^A_c,tau,W_c,e^A_c^top,tau dtau, \
V_d &= V_c,T,
endalign

where the subscripts $c$ and $d$ stand for the continuous and discrete time state space models respectively and $T$ the discretization time step size. The continuous time model would be

beginalign
dotx(t) &= A_c,x(t) + B_c,u(t) + w(t), quad w(t) sim mathcalN(0,W_c), \
y(t) &= C_c,x(t) + D_c,u(t) + v(t), quad v(t) sim mathcalN(0,V_c),
endalign

such that the discrete time model becomes

beginalign
x_k+1 &= A_d,x_k + B_d,u_k + w_k, quad w_k sim mathcalN(0,W_d), \
y_k &= C_d,x_k + D_d,u_k + v_k, quad v_k sim mathcalN(0,V_d).
endalign

When all matrices are constant during the discretization time step the expression for $W_d$ can be simplified to the following impliciet equation

$$
A_c,W_d + W_d,A_c^top = A_d,W_c,A_d^top - W_c.
$$

So if $A_d$, $A_c$ and $W_d$ are known the above equation is just a discrete Lyapunov equation in $W_c$.

share|cite|improve this answer

answered Mar 23 at 16:19

Kwin van der VeenKwin van der Veen

5,6502828

$endgroup$

share|cite|improve this answer

answered Mar 23 at 16:19

Kwin van der VeenKwin van der Veen

5,6502828

answered Mar 23 at 16:19

Kwin van der VeenKwin van der Veen

5,6502828

answered Mar 23 at 16:19

Kwin van der VeenKwin van der Veen

5,6502828